/* Easy
Given an array of characters, compress it in-place.
The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.
After you are done modifying the input array in-place, return the new length of the array.

Follow up:
Could you solve it using only O(1) extra space?

Example 1:
Input:
['a','a','b','b','c','c','c']

Output:
Return 6, and the first 6 characters of the input array should be: ['a','2','b','2','c','3']

Explanation:
'aa' is replaced by 'a2'. 'bb' is replaced by 'b2'. 'ccc' is replaced by 'c3'.
Example 2:
Input:
['a']

Output:
Return 1, and the first 1 characters of the input array should be: ['a']

Explanation:
Nothing is replaced.
Example 3:
Input:
['a','b','b','b','b','b','b','b','b','b','b','b','b']

Output:
Return 4, and the first 4 characters of the input array should be: ['a','b','1','2'].

Explanation:
Since the character 'a' does not repeat, it is not compressed. 'bbbbbbbbbbbb' is replaced by 'b12'.
Notice each digit has it's own entry in the array.

Note:
All characters have an ASCII value in [35, 126].
1 <= len(chars) <= 1000

Relatives:
038. Count and Say
271. Encode and Decode Strings
394. Decode String
443. String Compression
604. Design Compressed String Iterator
1313. Decompress Run-Length Encoded List */

#include <iostream>
#include <vector>
#include <stack>
#include <string>

using std::vector;

class Solution {
public:
    int compress(vector<char>& chars) {
        int i = 0;
        std::stack<int> stk;

        for (auto j = 0; j < chars.size();) {
            int cnt = 0;
            chars[i] = chars[j];

            while (j < chars.size() && chars[j] == chars[i]) {
                ++j;
                ++cnt;
            }
            if (cnt > 1) {
                for (const auto& chr : std::to_string(cnt)) {
                    chars[++i] = chr;
                }
            }
            ++i;
            // if (cnt > 1) {
            //     while (cnt) {
            //         stk.push(cnt % 10 + '0');
            //         cnt /= 10;
            //     }
            //     while (!stk.empty()) {
            //         chars[++i] = stk.top();
            //         stk.pop();
            //     }
            // }
        }

        return i;
    }
};